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$$\left(\sum_{i=1}^na_i^p\right)^{1/p} \ge \left(\sum_{i=1}^na_i^q\right)^{1/q} $$ if $0 < p \le q$ for $a_i\ge 0$. I have proved that the inequality holds for $ p=q $ (trivial) and I have also proved that it holds if one of the two sums is equal to 1, but I don't know how to continue. Please help.

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    $\begingroup$ Also, it is trivial if the sums are $0$. In all other cases, how do they behave under scaling (multiplying all $a_i$ by some constant $c$)? $\endgroup$ – Daniel Fischer Nov 2 '14 at 21:07
  • $\begingroup$ Have you heard about convexity? $\endgroup$ – Pedro Tamaroff Nov 2 '14 at 21:08
  • $\begingroup$ no man i have not heard about it $\endgroup$ – Baquesh Nov 2 '14 at 21:09
  • $\begingroup$ Prove Jensen's inequality - Why ? Has Jensen ever proved one of mine ? Course he hasn't ! And he expects me to prove one of his now, does he !? $\endgroup$ – Lucian Nov 2 '14 at 21:14
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You want to prove that, for $0 < p \leq q <\infty$, $$ \left(\sum_i \lvert{x_i}\rvert^q\right)^{1/q} = \lVert{x}\rVert_q \leq \lVert{x}\rVert_p = \left(\sum_i \lvert{x_i}\rvert^p\right)^{1/p}\;. $$ To see why, one can easily prove that if $\lVert{x}\rVert_p = 1$, then $\lVert{x}\rVert_q^q \leq 1$ (bounding each term $\lvert{x_i}\rvert^q \leq \lvert{x_i}\rvert^p$), and therefore $\lVert{x}\rVert_q \leq 1 = \lVert{x}\rVert_p$. Next, for the general case, apply this to $y = x/\lVert{x}\rVert_p$, which has unit $\lVert{\cdot}\rVert_p$ norm, and conclude by homogeneity of the norm.

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  • $\begingroup$ these are all new terms to me could you explain more clearly please $\endgroup$ – Baquesh Nov 2 '14 at 21:23
  • $\begingroup$ Replace my $x$'s by your $a$'s, and you obtain your question. What is confusing you? $\endgroup$ – Clement C. Nov 2 '14 at 21:33

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